Solution

Since the volume of the vessel remains constant, $V_1 = V_2 = V$.

Using the ideal gas law, the ratio of pressure and volume to mass and temperature is constant for a given gas:
$$\frac{p_1 V}{m_1 T_1} = \frac{p_2 V}{m_2 T_2}$$

Expressing the final pressure $p_2$:
$$p_2 = p_1 \frac{T_2}{T_1} \frac{m_2}{m_1}$$

We are given the initial pressure $p_1 = 2 \, \text{MPa} = 2 \cdot 10^6 \, \text{Pa}$.
The initial temperature is $T_1 = 27 \, ^\circ\text{C} = 300 \, \text{K}$.
The gas is heated by $50 \, ^\circ\text{C}$, so the final temperature is $T_2 = 300 + 50 = 350 \, \text{K}$.
The remaining mass is half of the initial mass, so $m_2 = \frac{m_1}{2}$.

Substituting these into the equation:
$$p_2 = \frac{p_1 T_2}{2 T_1}$$

Calculating the value:
$$p_2 = \frac{2 \cdot 10^6 \cdot 350}{2 \cdot 300} \approx 1.166 \cdot 10^6 \, \text{Pa} = 1.166 \, \text{MPa}$$